Use rolling-ball algorithm for estimating background intensity

The rolling-ball algorithm estimates the background intensity of a grayscale image in case of uneven exposure. It is frequently used in biomedical image processing and was first proposed by Stanley R. Sternberg in 1983 1.

The algorithm works as a filter and is quite intuitive. We think of the image as a surface that has unit-sized blocks stacked on top of each other in place of each pixel. The number of blocks, and hence surface height, is determined by the intensity of the pixel. To get the intensity of the background at a desired (pixel) position, we imagine submerging a ball under the surface at the desired position. Once it is completely covered by the blocks, the apex of the ball determines the intensity of the background at that position. We can then roll this ball around below the surface to get the background values for the entire image.

Scikit-image implements a general version of this rolling-ball algorithm, which allows you to not just use balls, but arbitrary shapes as kernel and works on n-dimensional ndimages. This allows you to directly filter RGB images or filter image stacks along any (or all) spacial dimensions.

1

Sternberg, Stanley R. “Biomedical image processing.” Computer 1 (1983): 22-34. DOI:10.1109/MC.1983.1654163

Classic rolling ball

In scikit-image, the rolling ball algorithm assumes that your background has low intensity (black), whereas the features have high intensity (white). If this is the case for your image, you can directly use the filter like so:

import imageio
import matplotlib.pyplot as plt
import numpy as np

from skimage import (
    data, restoration, util
)


def plot_result(image, background):
    fig, ax = plt.subplots(nrows=1, ncols=3)

    ax[0].imshow(image, cmap='gray')
    ax[0].set_title('Original image')
    ax[0].axis('off')

    ax[1].imshow(background, cmap='gray')
    ax[1].set_title('Background')
    ax[1].axis('off')

    ax[2].imshow(image - background, cmap='gray')
    ax[2].set_title('Result')
    ax[2].axis('off')

    fig.tight_layout()


image = data.coins()

background = restoration.rolling_ball(image)

plot_result(image, background)
plt.show()
Original image, Background, Result

White background

If you have dark features on a bright background, you need to invert the image before you pass it into the algorithm, and then invert the result. This can be accomplished via:

image = data.page()
image_inverted = util.invert(image)

background_inverted = restoration.rolling_ball(image_inverted, radius=45)
filtered_image_inverted = image_inverted - background_inverted
filtered_image = util.invert(filtered_image_inverted)
background = util.invert(background_inverted)

fig, ax = plt.subplots(nrows=1, ncols=3)

ax[0].imshow(image, cmap='gray')
ax[0].set_title('Original image')
ax[0].axis('off')

ax[1].imshow(background, cmap='gray')
ax[1].set_title('Background')
ax[1].axis('off')

ax[2].imshow(filtered_image, cmap='gray')
ax[2].set_title('Result')
ax[2].axis('off')

fig.tight_layout()

plt.show()
Original image, Background, Result

Be careful not to fall victim to an integer underflow when subtracting a bright background. For example, this code looks correct, but may suffer from an underflow leading to unwanted artifacts. You can see this in the top right corner of the visualization.

image = data.page()
image_inverted = util.invert(image)

background_inverted = restoration.rolling_ball(image_inverted, radius=45)
background = util.invert(background_inverted)
underflow_image = image - background  # integer underflow occurs here

# correct subtraction
correct_image = util.invert(image_inverted - background_inverted)

fig, ax = plt.subplots(nrows=1, ncols=2)

ax[0].imshow(underflow_image, cmap='gray')
ax[0].set_title('Background Removal with Underflow')
ax[0].axis('off')

ax[1].imshow(correct_image, cmap='gray')
ax[1].set_title('Correct Background Removal')
ax[1].axis('off')

fig.tight_layout()

plt.show()
Background Removal with Underflow, Correct Background Removal

Image Datatypes

rolling_ball can handle datatypes other than np.uint8. You can pass them into the function in the same way.

image = data.coins()[:200, :200].astype(np.uint16)

background = restoration.rolling_ball(image, radius=70.5)
plot_result(image, background)
plt.show()
Original image, Background, Result

However, you need to be careful if you use floating point images that have been normalized to [0, 1]. In this case the ball will be much larger than the image intensity, which can lead to unexpected results.

Original image, Background, Result

Because radius=70.5 is much larger than the maximum intensity of the image, the effective kernel size is reduced significantly, i.e., only a small cap (approximately radius=10) of the ball is rolled around in the image. You can find a reproduction of this strange effect in the Advanced Shapes section below.

To get the expected result, you need to reduce the intensity of the kernel. This is done by specifying the kernel manually using the kernel argument.

Note: The radius is equal to the length of a semi-axis of an ellipsis, which is half a full axis. Hence, the kernel shape is multipled by two.

Original image, Background, Result

Advanced Shapes

By default, rolling_ball uses a ball shaped kernel (surprise). Sometimes, this can be too limiting - as in the example above -, because the intensity dimension has a different scale compared to the spatial dimensions, or because the image dimensions may have different meanings - one could be a stack counter in an image stack.

To account for this, rolling_ball has a kernel argument which allows you to specify the kernel to be used. A kernel must have the same dimensionality as the image (Note: dimensionality, not shape). To help with it’s creation, two default kernels are provided by skimage. ball_kernel specifies a ball shaped kernel and is used as the default kernel. ellipsoid_kernel specifies an ellipsoid shaped kernel.

Original image, Background, Result

You can also use ellipsoid_kernel to recreate the previous, unexpected result and see that the effective (spatial) filter size was reduced.

Original image, Background, Result

Higher Dimensions

Another feature of rolling_ball is that you can directly apply it to higher dimensional images, e.g., a z-stack of images obtained during confocal microscopy. The number of kernel dimensions must match the image dimensions, hence the kernel shape is now 3 dimensional.

image = data.cells3d()[:, 1, ...]
background = restoration.rolling_ball(
    image,
    kernel=restoration.ellipsoid_kernel(
        (1, 21, 21),
        0.1
    )
)

plot_result(image[30, ...], background[30, ...])
plt.show()
Original image, Background, Result

A kernel size of 1 does not filter along this axis. In other words, above filter is applied to each image in the stack individually.

However, you can also filter along all 3 dimensions at the same time by specifying a value other than 1.

image = data.cells3d()[:, 1, ...]
background = restoration.rolling_ball(
    image,
    kernel=restoration.ellipsoid_kernel(
        (5, 21, 21),
        0.1
    )
)

plot_result(image[30, ...], background[30, ...])
plt.show()
Original image, Background, Result

Another possibility is to filter individual pixels along the planar axis (z-stack axis).

image = data.cells3d()[:, 1, ...]
background = restoration.rolling_ball(
    image,
    kernel=restoration.ellipsoid_kernel(
        (100, 1, 1),
        0.1
    )
)

plot_result(image[30, ...], background[30, ...])
plt.show()
Original image, Background, Result

Total running time of the script: ( 1 minutes 29.634 seconds)

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